0
\$\begingroup\$

Given a fixed space on a solar panel, all else equal, adding multiple cells in series increases the voltage (while current remains the same). For example, two cells equals 0.5v + 0.5v = 1v total output. Three cells equals 0.5v + 0.5v + 0.5v = 1.5v, etc.

My question is, why wouldn't solar panel manufacturers create infinitely small cells (that are still workable) in parallel in order to maximize voltage? For example, on a solar panel, wouldn't reducing the size of solar cells to half the size (effectively doubling the number of cells on the panel while keeping the same area) double the voltage?

\$\endgroup\$
3
  • \$\begingroup\$ While solar cells are still slabs of silicon, the fewer things you have to solder together, the better. \$\endgroup\$
    – user16324
    Commented Jul 7, 2020 at 18:30
  • \$\begingroup\$ reducing the size of solar cells to half the size ... double the voltage? Yes but that would also halve the current. As Power = Voltage * Current, the amount of power would remain the same. \$\endgroup\$ Commented Jul 7, 2020 at 18:38
  • \$\begingroup\$ At what point? 1 cell, at one cell adding another in parallel will not increase the voltage. \$\endgroup\$
    – MadHatter
    Commented Jul 7, 2020 at 23:40

2 Answers 2

2
\$\begingroup\$

If you are referring to a solar panel rather than a complete installation then doubling the number of cells would indeed get twice the amount of voltage but each cell would only be able to provide half the current so this will not increase the total amount of power available from the panel.

Since there is lost space at the edge of each cell you won't be able to get quite the same amount of area of cells so the power will be slightly less and it will cost more because of increased number of connections and manufacturing steps.

The voltage outputs of photovoltaic panels have traditionally standardized on a few voltages, in particular, voltage suitable for use with 12v lead-acid batteries so most arrays are designed around that. This will require about 32-36 cells in each panel to provide that voltage. The open-circuit voltage will typically be in the range of 18-21 volts with a maximum power voltage of ~12-15v - the same as a lead-acid battery.

For an entire array made up of many individual panels they are typically connected in series to give an operating voltage that can be up to several hundred volts. Beyond that the expense and difficulty of making the system safe creates a limit.

\$\endgroup\$
0
\$\begingroup\$

Splitting a cell in half into series elements doubles the voltage and almost halves the current and thus Impedance at MPT is 4 x N splits yet power is constant in theory until the gaps and connections reduce the useable space of a cell for each split.

The optimum cell array of X series and Y parallel cells perhaps noted as xSyP will also affect the cell impedance Ra=Rc X/Y.

Since maximum power is always desired, the impedance and voltage must be matched. This requires choosing the array size to conveniently match up to popular battery charging systems.

Cell splitting to infinite would yield too many interconnections with very high impedance, But to determine the optimum where power conversion would be practical relies on knowing the cost of impedance and voltage on all the parts involved. On FETs, for example, the cost is inversely proportional to RdsOn but then has some voltage limit that is hard to exceed (? kV) so there is a cost penalty for insulation gaps between conductors. These may be one factor.

The result in large kW or MW PV panels the working voltage will rise above 500V. I do not know how they are defined, but the cost of total ownership will be the overall driver for these standards.

In small arrays, they MPT voltage at max power is usually around 82% of open-circuit voltage, and 70% for 10% Max solar input so the Voc cell array is chosen to match popular battery voltages and may be slightly higher the float voltage for simple chargers. But remember the cell is a solar controlled current source but has a source impedance = V/I at MPT Zmp=Vmp/Imp which is approximately near Zpv = Voc/Isc.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.